$h(n) = -n^{2}+2n-2(f(n))$ $f(t) = 6t^{3}-3t^{2}-6t$ $g(n) = -6n-5(f(n))$ $ f(h(-1)) = {?} $
First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = -(-1)^{2}+(2)(-1)-2(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = 6(-1)^{3}-3(-1)^{2}+(-6)(-1)$ $f(-1) = -3$ That means $h(-1) = -(-1)^{2}+(2)(-1)+(-2)(-3)$ $h(-1) = 3$ Now we know that $h(-1) = 3$ . Let's solve for $f(h(-1))$ , which is $f(3)$ $f(3) = 6(3^{3})-3(3^{2})+(-6)(3)$ $f(3) = 117$